The equation of a circle is $$ (x+2)^{2}+(y-3)^{2}=9 $$. What are the coordinates of the center of this circle?

Question

UpStudy AI · Accepted Answer

To find the coordinates of the center of the circle given by the equation

$$
(x+2)^{2}+(y-3)^{2}=9,
$$

we can identify the center from the standard form of the equation of a circle, which is

$$
(x-h)^{2}+(y-k)^{2}=r^{2},
$$

where $$(h, k)$$ is the center of the circle and $$r$$ is the radius.

From the given equation:

- The term $$(x+2)^{2}$$ indicates that $$h = -2$$ (since it is in the form $$(x - h)$$).
- The term $$(y-3)^{2}$$ indicates that $$k = 3$$.

Thus, the coordinates of the center of the circle are

$$
(-2, 3).
$$

Therefore, the center of the circle is at the point $$(-2, 3)$$.
The center of the circle is at $$(-2, 3)$$.

The equation of a circle is . What are the
coordinates of the center of this circle?